Assume first that the teams are labelled, with the two teams of three having labels A and B, and the three teams of four having labels C, D, and E.
There are $\binom{18}{3}$ ways to choose the people who will be on Team A. For each of these ways, there are $\binom{15}{3}$ ways to choose Team B. For every way to choose Teams A and B, there are $\binom{12}{4}$ ways to choose Team C, and then $\binom{8}{4}$ ways to choose Team D, and finally, if we like, $\binom{4}{4}$ ways to choose Team E.
However, when we remove the labels, the number of choices for the $2$ three-person teams gets divided by $2!$, and the number of choices for the $3$ four-person teams gets divided by $3!$, for a total of
$$\frac{\binom{18}{3}\binom{15}{3}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{2!3!}.$$
For the Mia and Max problem, we could count. But we prefer to work directly with probabilities.
Imagine that the people first get divided into $2$ groups, of $6$ and $12$, to make up the two kinds of team.
The probability that Mia is selected for the group of $6$ is $\frac{6}{18}$. Given that this has happened, the probability Max is chosen for the same team is $\frac{2}{17}$. Thus the probability Mia and Max are in the same team of three is
$$\frac{6}{18}\cdot\frac{2}{17}.$$
Similarly, the probability that Mia and Max are in the same group of $4$ is
$$\frac{12}{18}\cdot\frac{3}{17}.$$
Add.
Remark: Doing problem (b) by counting and dividing is not difficult, just a little more messy-looking. Note that the probabilities are the same for the labelled teams case as for the unlabelled case. It is easier not to make a mistake by counting the number of ways that Mia and Max can be on the same labelled team, and dividing by the number of labelled teams.
Best Answer
John is in one class, so there are 29 remaining slots in his class and 30 slots in the other class. Jane has a $29/(29+30) = 29/59$ chance of being in John's class.